keras (tensorflow backend) conditional assignment with K.switch() - if-statement

I'm trying to implement something like
if np.max(subgrid) == np.min(subgrid):
middle_middle = cur_subgrid + 1
else:
middle_middle = cur_subgrid
Since the condition can only be determined at run-time, I'm using Keras syntax as following
middle_middle = K.switch(K.max(subgrid) == K.min(subgrid), lambda: tf.add(cur_subgrid,1), lambda: cur_subgrid)
But I'm getting this error:
<ipython-input-112-0504ce070e71> in col_loop(j, gray_map, mask_A)
56
57
---> 58 middle_middle = K.switch(K.max(subgrid) == K.min(subgrid), lambda: tf.add(cur_subgrid,1), lambda: cur_subgrid)
59
60 print ('ml',middle_left.shape)
/nfs/isicvlnas01/share/anaconda3/lib/python3.5/site-packages/keras/backend/tensorflow_backend.py in switch(condition, then_expression, else_expression) 2561 The selected tensor. 2562 """
-> 2563 if condition.dtype != tf.bool: 2564 condition = tf.cast(condition, 'bool') 2565 if not callable(then_expression):
AttributeError: 'bool' object has no attribute 'dtype'
middle_middle, cur_subgrid, and subgrid are all NxN tensors. Any help is appreciated.

I think the problem is that with K.max(subgrid) == K.min(subgrid) you're creating a python boolean comparing two tensor objects, not a tensorflow boolean tensor containing the value of the comparison of the values of the two input tensors.
In other words, what you have written will be evaluated as
K.switch(False, lambda: tf.add(cur_subgrid,1), lambda: cur_subgrid)
instead of
comparison = ... # Some tensor, that at runtime will contain True if min and max are the same, False otherwise.
K.switch(comparison , lambda: tf.add(cur_subgrid,1), lambda: cur_subgrid)
So what you need to do is to use keras.backend.equal() instead of ==:
K.switch(K.equal(K.max(subgrid),K.min(subgrid)), lambda: tf.add(cur_subgrid,1), lambda: cur_subgrid)

Related

Creating efficiently a dictionary in pyspark using list derived from df columns

I created this function in python using pandas dataframe, and I'd like to use it also in spark.
What I'm doing with this function is :
converting the df column to a list ( t1 )
converting the unique values of the column to a list ( t2 )
creating a list for each unique value of each feature ( t ). this list takes value 1 when the unique value is present in t1, 0 otherwise.
at the end the result is a dictionary with the unique values of each feature as key and as argument a list with value 1 when the key (the unique value) appears and 0 otherwise.
feat_list is just a list with all the column names.
def binary_dict(pandas_df, feat_list):
dict_feature = dict()
for col in feat_list:
t1 = pandas_df[col].tolist()
t2 = pandas_df[col].unique().tolist()
for value in t2:
t = []
for i in range (0, len(t1)):
if value == t1[i]:
t.append(1)
else:
t.append(0)
cc = str(col)
vv = "_" + str(value)
cv = cc + vv
dict_feature[cv] = t
return dict_feature
I tried using
t1 = df.select("col_name").rdd.flatMap(list).collect()
for creating t1 but it took over 20 minutes to create the list for a single column. I got something like 100 columns. Is there a way to convert this function to spark efficiently?
Thanks everyone for the answers!
PS: I'm using synapse analytics by azure/microsoft, Python 3.8 and pyspark 3.1.

How can I convert +00:00 timezone to Z timezone

I am comparing the following dates as:
result = 2018-06-29T20:56:41+00:00 <= 2018-06-30T00:38:32Z
But this is giving as false. How can I make the 2 dates compare to True as Cleary 29 < 30. Initially, I thought it has to do with the timezone but on google search, found out that both formats are UTC timezone. Can anyone help me understand if that's correct and then compare these results to true?
Are you actually converting them to datatime objects, e.g.:
In []:
d1 = datetime.strptime("2018-06-29T20:56:41+00:00", "%Y-%m-%dT%H:%M:%S%z")
d2 = datetime.strptime("2018-06-30T00:38:32Z", "%Y-%m-%dT%H:%M:%S%z")
d1 <= d2
Out[]:
True
Note: in Py3.7 the first of these could be replaced with datetime.fromisoformat()
But even the string forms should also return True, so not sure why you are getting False:
In []:
"2018-06-29T20:56:41+00:00" <= "2018-06-30T00:38:32Z"
Out[]:
True

Reuse data in a Tensorflow graph (using Queues and tf.cond())

I'm building a resample layer in Tensorflow that is meant to reuse data for resample_n-times before getting new data from the net_data_layer() (a Que reading from some hdf5 datasources). I have a Network class and store the tensors returned from the data layer in it (self) for "global" access (e.g. for building the network-graph, but also by my train_op()):
self.batch_img, self.batch_label, self.batch_weights = net_data_layer()
I added some logic so that in the first iteration resample == False, with the intention of data-layer queues being called, and then resample == True for the next resample_n-th iterations. The part in the graph boils down to:
self.batch_img, self.batch_label, self.batch_weights = \
tf.cond(resample,
lambda: resample_data(),
lambda: net_data_layer())
Where resample_data() just returns the previously stored data, basically doing nothing but forwarding the already stored data to the graph:
def resample_data(): return self.batch_img, self.batch_label, self.batch_weights
But I'm getting an error:
ValueError: Operation 'ResampleLayer/cond/DataLayer_hdf5_dset/test_data/fifo_queue_enqueue' has been marked as not fetchable.
This led me to this discussion, where this was solved by "Putting the QueueRunner logic outside the tf.cond() context". However having the Queues IN there is the whole point of my condition, since when I run them (specifically the dequeue-op), they jump to the next sample in line.
Is there at all a way to make the graph reuse its already loaded data (or would that be kind of "cyclic" and isn't supported?). I'm out of depth... any ideas?
For reference the whole "ResampleLayer" code block:
with tf.variable_scope('ResampleLayer'):
resample_n = args.resample_n # user-defined
resample_i = 0 # resample counter, when 0 it generates new data, then counts up to resample_n
# dummy placeholders, needed for first run since tf.cond() doesn't do lazy evaluation
self.batch_img = tf.constant(0., dtype=tf.float32, shape=[batch_size]+shape_img)
self.batch_label = tf.constant(0., dtype=tf.float32, shape=[batch_size]+shape_label)
self.batch_weights = tf.constant(0., dtype=tf.float32, shape=[batch_size]+shape_weights)
def net_data_layer():
# run the data layer to generate a new batch
self.batch_img, self.batch_label, self.batch_weights = \
data_layer.data_TFRqueue(dataset_pth, batch_size=batch_size)
return self.batch_img, self.batch_label, self.batch_weights
def resample_data():
# just forward the existing batch
return self.batch_img, self.batch_label, self.batch_weights
with tf.variable_scope('ResampleLogic'):
t_resample_n = tf.Variable(self.resample_n, tf.int8)
t_resample_i = tf.Variable(self.resample_i, tf.int8)
# Determine when to resample. Pure Python equivalent:
# resample = False if resample_i == 0 else True
resample = tf.cond(tf.equal(t_resample_i, tf.constant(0)), lambda: tf.constant(False), lambda: tf.constant(True))
# "Loop" the resample counter. Pure Python equivalent:
# resample_i = resample_i + 1 if resample_i < resample_n else 0
tf.cond(t_resample_i < t_resample_n, lambda: t_resample_i + 1, lambda: tf.constant(0))
# the actual intended branching of the graph
self.batch_img, self.batch_label, self.batch_weights = \
tf.cond(resample,
lambda: resample_data(),
lambda: net_data_layer())
## NETWORK
self.output_mask = self._build_net(self.batch_img, ...)

Error when using ismatch() function with regex in Julia

I'm trying to do a very simple program to find matches with ismatch() function in Julia. Suppose my pattern is
e_pat = r".+#.+"
Then I make a list called input with some random elements:
input= ["pipo#gmail.com", 23, "trapo", "holi#gmail.com"]
Now I want to identify how many matches exist and then print them using e_pat as reference:
for i in input
println(ismatch(e_pat, i)) && println(i)
end
With that code I just get "true" and the error displayed below:
true
TypeError: non-boolean (Void) used in boolean context
Stacktrace:
[1] macro expansion at ./In[27]:4 [inlined]
[2] anonymous at ./<missing>:?
[3] include_string(::String, ::String) at ./loading.jl:522
What can I do in order to get the following?
"pipo#gmail.com"
"holi#gmail.com"
I read ismatch() documentation but found nothing useful.
Any help will be much appreciated
The problem is that while this expression returns true:
julia> #show ismatch(e_pat, "pipo#gmail.com");
ismatch(e_pat,"pipo#gmail.com") = true
Using println, just prints true but returns nothing:
julia> #show println(ismatch(e_pat, "pipo#gmail.com"));
true
println(ismatch(e_pat,"pipo#gmail.com")) = nothing
Which is of type Void:
julia> typeof(nothing)
Void
And the error is telling you that you cant use an object of type Void in a boolean context (nothing) is just an instance of Void treated like a singleton in Julia:
julia> nothing && true
ERROR: TypeError: non-boolean (Void) used in boolean context
After fixing that also notice that this is also another error:
julia> #show ismatch(e_pat, 42);
ERROR: MethodError: no method matching ismatch(::Regex, ::Int32)
Closest candidates are:
ismatch(::Regex, ::SubString{T<:AbstractString}) at regex.jl:151
ismatch(::Regex, ::SubString{T<:AbstractString}, ::Integer) at regex.jl:151
ismatch(::Regex, ::AbstractString) at regex.jl:145
...
This is telling you that ismatch has no such method, you can't use it with a combination of arguments of types: (Regex, Int).
You could do something like this instead to make sure all the objects are Strings:
julia> input = string.(["pipo#gmail.com", 23, "trapo", "holi#gmail.com"])
4-element Array{String,1}:
"pipo#gmail.com"
"23"
"trapo"
"holi#gmail.com"
Finally, you could use the macro #show (which prints an expression and its result and finally returns the result) instead of the println function (which prints the result and returns nothing, to debug whats going on:
julia> for i in input
#show(ismatch(e_pat, i)) && println(i)
end
ismatch(e_pat,i) = true
pipo#gmail.com
ismatch(e_pat,i) = false
ismatch(e_pat,i) = false
ismatch(e_pat,i) = true
holi#gmail.com
So in order to print your expected result just remove the left hand side println:
julia> for i in input
ismatch(e_pat, i) && println(i)
end
pipo#gmail.com
holi#gmail.com
If you want to store them instead of printing them you could us an array comprehension instead:
julia> result = [str for str in input if ismatch(e_pat, str)]
2-element Array{String,1}:
"pipo#gmail.com"
"holi#gmail.com"
Or an indexing expression like this one:
julia> ismatch.(e_pat, input)
4-element BitArray{1}:
true
false
false
true
julia> result = input[ismatch.(e_pat, input)]
2-element Array{String,1}:
"pipo#gmail.com"
"holi#gmail.com"
That way you could print them later without having to repeat the computation:
julia> println.(result)
pipo#gmail.com
holi#gmail.com

Python consecutive if statements [duplicate]

Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
Is there a ternary conditional operator in Python?
Yes, it was added in version 2.5. The expression syntax is:
a if condition else b
First condition is evaluated, then exactly one of either a or b is evaluated and returned based on the Boolean value of condition. If condition evaluates to True, then a is evaluated and returned but b is ignored, or else when b is evaluated and returned but a is ignored.
This allows short-circuiting because when condition is true only a is evaluated and b is not evaluated at all, but when condition is false only b is evaluated and a is not evaluated at all.
For example:
>>> 'true' if True else 'false'
'true'
>>> 'true' if False else 'false'
'false'
Note that conditionals are an expression, not a statement. This means you can't use statements such as pass, or assignments with = (or "augmented" assignments like +=), within a conditional expression:
>>> pass if False else pass
File "<stdin>", line 1
pass if False else pass
^
SyntaxError: invalid syntax
>>> # Python parses this as `x = (1 if False else y) = 2`
>>> # The `(1 if False else x)` part is actually valid, but
>>> # it can't be on the left-hand side of `=`.
>>> x = 1 if False else y = 2
File "<stdin>", line 1
SyntaxError: cannot assign to conditional expression
>>> # If we parenthesize it instead...
>>> (x = 1) if False else (y = 2)
File "<stdin>", line 1
(x = 1) if False else (y = 2)
^
SyntaxError: invalid syntax
(In 3.8 and above, the := "walrus" operator allows simple assignment of values as an expression, which is then compatible with this syntax. But please don't write code like that; it will quickly become very difficult to understand.)
Similarly, because it is an expression, the else part is mandatory:
# Invalid syntax: we didn't specify what the value should be if the
# condition isn't met. It doesn't matter if we can verify that
# ahead of time.
a if True
You can, however, use conditional expressions to assign a variable like so:
x = a if True else b
Or for example to return a value:
# Of course we should just use the standard library `max`;
# this is just for demonstration purposes.
def my_max(a, b):
return a if a > b else b
Think of the conditional expression as switching between two values. We can use it when we are in a 'one value or another' situation, where we will do the same thing with the result, regardless of whether the condition is met. We use the expression to compute the value, and then do something with it. If you need to do something different depending on the condition, then use a normal if statement instead.
Keep in mind that it's frowned upon by some Pythonistas for several reasons:
The order of the arguments is different from those of the classic condition ? a : b ternary operator from many other languages (such as C, C++, Go, Perl, Ruby, Java, JavaScript, etc.), which may lead to bugs when people unfamiliar with Python's "surprising" behaviour use it (they may reverse the argument order).
Some find it "unwieldy", since it goes contrary to the normal flow of thought (thinking of the condition first and then the effects).
Stylistic reasons. (Although the 'inline if' can be really useful, and make your script more concise, it really does complicate your code)
If you're having trouble remembering the order, then remember that when read aloud, you (almost) say what you mean. For example, x = 4 if b > 8 else 9 is read aloud as x will be 4 if b is greater than 8 otherwise 9.
Official documentation:
Conditional expressions
Is there an equivalent of C’s ”?:” ternary operator?
You can index into a tuple:
(falseValue, trueValue)[test]
test needs to return True or False.
It might be safer to always implement it as:
(falseValue, trueValue)[test == True]
or you can use the built-in bool() to assure a Boolean value:
(falseValue, trueValue)[bool(<expression>)]
For versions prior to 2.5, there's the trick:
[expression] and [on_true] or [on_false]
It can give wrong results when on_true has a false Boolean value.1
Although it does have the benefit of evaluating expressions left to right, which is clearer in my opinion.
1. Is there an equivalent of C’s ”?:” ternary operator?
<expression 1> if <condition> else <expression 2>
a = 1
b = 2
1 if a > b else -1
# Output is -1
1 if a > b else -1 if a < b else 0
# Output is -1
From the documentation:
Conditional expressions (sometimes called a “ternary operator”) have the lowest priority of all Python operations.
The expression x if C else y first evaluates the condition, C (not x); if C is true, x is evaluated and its value is returned; otherwise, y is evaluated and its value is returned.
See PEP 308 for more details about conditional expressions.
New since version 2.5.
An operator for a conditional expression in Python was added in 2006 as part of Python Enhancement Proposal 308. Its form differs from common ?: operator and it looks like this:
<expression1> if <condition> else <expression2>
which is equivalent to:
if <condition>: <expression1> else: <expression2>
Here is an example:
result = x if a > b else y
Another syntax which can be used (compatible with versions before 2.5):
result = (lambda:y, lambda:x)[a > b]()
where operands are lazily evaluated.
Another way is by indexing a tuple (which isn't consistent with the conditional operator of most other languages):
result = (y, x)[a > b]
or explicitly constructed dictionary:
result = {True: x, False: y}[a > b]
Another (less reliable), but simpler method is to use and and or operators:
result = (a > b) and x or y
however this won't work if x would be False.
A possible workaround is to make x and y lists or tuples as in the following:
result = ((a > b) and [x] or [y])[0]
or:
result = ((a > b) and (x,) or (y,))[0]
If you're working with dictionaries, instead of using a ternary conditional, you can take advantage of get(key, default), for example:
shell = os.environ.get('SHELL', "/bin/sh")
Source: ?: in Python at Wikipedia
Unfortunately, the
(falseValue, trueValue)[test]
solution doesn't have short-circuit behaviour; thus both falseValue and trueValue are evaluated regardless of the condition. This could be suboptimal or even buggy (i.e. both trueValue and falseValue could be methods and have side effects).
One solution to this would be
(lambda: falseValue, lambda: trueValue)[test]()
(execution delayed until the winner is known ;)), but it introduces inconsistency between callable and non-callable objects. In addition, it doesn't solve the case when using properties.
And so the story goes - choosing between three mentioned solutions is a trade-off between having the short-circuit feature, using at least Python 2.5 (IMHO, not a problem anymore) and not being prone to "trueValue-evaluates-to-false" errors.
Ternary operator in different programming languages
Here I just try to show some important differences in the ternary operator between a couple of programming languages.
Ternary operator in JavaScript
var a = true ? 1 : 0;
# 1
var b = false ? 1 : 0;
# 0
Ternary operator in Ruby
a = true ? 1 : 0
# 1
b = false ? 1 : 0
# 0
Ternary operator in Scala
val a = true ? 1 | 0
# 1
val b = false ? 1 | 0
# 0
Ternary operator in R programming
a <- if (TRUE) 1 else 0
# 1
b <- if (FALSE) 1 else 0
# 0
Ternary operator in Python
a = 1 if True else 0
# 1
b = 1 if False else 0
# 0
For Python 2.5 and newer there is a specific syntax:
[on_true] if [cond] else [on_false]
In older Pythons, a ternary operator is not implemented but it's possible to simulate it.
cond and on_true or on_false
Though there is a potential problem, which is if cond evaluates to True and on_true evaluates to False then on_false is returned instead of on_true. If you want this behaviour the method is OK, otherwise use this:
{True: on_true, False: on_false}[cond is True] # is True, not == True
which can be wrapped by:
def q(cond, on_true, on_false)
return {True: on_true, False: on_false}[cond is True]
and used this way:
q(cond, on_true, on_false)
It is compatible with all Python versions.
You might often find
cond and on_true or on_false
but this leads to a problem when on_true == 0
>>> x = 0
>>> print x == 0 and 0 or 1
1
>>> x = 1
>>> print x == 0 and 0 or 1
1
Where you would expect this result for a normal ternary operator:
>>> x = 0
>>> print 0 if x == 0 else 1
0
>>> x = 1
>>> print 0 if x == 0 else 1
1
Does Python have a ternary conditional operator?
Yes. From the grammar file:
test: or_test ['if' or_test 'else' test] | lambdef
The part of interest is:
or_test ['if' or_test 'else' test]
So, a ternary conditional operation is of the form:
expression1 if expression2 else expression3
expression3 will be lazily evaluated (that is, evaluated only if expression2 is false in a boolean context). And because of the recursive definition, you can chain them indefinitely (though it may considered bad style.)
expression1 if expression2 else expression3 if expression4 else expression5 # and so on
A note on usage:
Note that every if must be followed with an else. People learning list comprehensions and generator expressions may find this to be a difficult lesson to learn - the following will not work, as Python expects a third expression for an else:
[expression1 if expression2 for element in iterable]
# ^-- need an else here
which raises a SyntaxError: invalid syntax.
So the above is either an incomplete piece of logic (perhaps the user expects a no-op in the false condition) or what may be intended is to use expression2 as a filter - notes that the following is legal Python:
[expression1 for element in iterable if expression2]
expression2 works as a filter for the list comprehension, and is not a ternary conditional operator.
Alternative syntax for a more narrow case:
You may find it somewhat painful to write the following:
expression1 if expression1 else expression2
expression1 will have to be evaluated twice with the above usage. It can limit redundancy if it is simply a local variable. However, a common and performant Pythonic idiom for this use-case is to use or's shortcutting behavior:
expression1 or expression2
which is equivalent in semantics. Note that some style-guides may limit this usage on the grounds of clarity - it does pack a lot of meaning into very little syntax.
One of the alternatives to Python's conditional expression
"yes" if boolean else "no"
is the following:
{True: "yes", False: "no"}[boolean]
which has the following nice extension:
{True: "yes", False: "no", None: "maybe"}[boolean_or_none]
The shortest alternative remains
("no", "yes")[boolean]
which works because issubclass(bool, int).
Careful, though: the alternative to
yes() if boolean else no()
is not
(no(), yes())[boolean] # bad: BOTH no() and yes() are called
but
(no, yes)[boolean]()
This works fine as long as no and yes are to be called with exactly the same parameters. If they are not, like in
yes("ok") if boolean else no() # (1)
or in
yes("ok") if boolean else no("sorry") # (2)
then a similar alternative either does not exist (1) or is hardly viable (2). (In rare cases, depending on the context, something like
msg = ("sorry", "ok")[boolean]
(no, yes)[boolean](msg)
could make sense.)
Thanks to Radek Rojík for his comment
As already answered, yes, there is a ternary operator in Python:
<expression 1> if <condition> else <expression 2>
In many cases <expression 1> is also used as Boolean evaluated <condition>. Then you can use short-circuit evaluation.
a = 0
b = 1
# Instead of this:
x = a if a else b
# Evaluates as 'a if bool(a) else b'
# You could use short-circuit evaluation:
x = a or b
One big pro of short-circuit evaluation is the possibility of chaining more than two expressions:
x = a or b or c or d or e
When working with functions it is more different in detail:
# Evaluating functions:
def foo(x):
print('foo executed')
return x
def bar(y):
print('bar executed')
return y
def blubb(z):
print('blubb executed')
return z
# Ternary Operator expression 1 equals to False
print(foo(0) if foo(0) else bar(1))
''' foo and bar are executed once
foo executed
bar executed
1
'''
# Ternary Operator expression 1 equals to True
print(foo(2) if foo(2) else bar(3))
''' foo is executed twice!
foo executed
foo executed
2
'''
# Short-circuit evaluation second equals to True
print(foo(0) or bar(1) or blubb(2))
''' blubb is not executed
foo executed
bar executed
1
'''
# Short-circuit evaluation third equals to True
print(foo(0) or bar(0) or blubb(2))
'''
foo executed
bar executed
blubb executed
2
'''
# Short-circuit evaluation all equal to False
print(foo(0) or bar(0) or blubb(0))
''' Result is 0 (from blubb(0)) because no value equals to True
foo executed
bar executed
blubb executed
0
'''
PS: Of course, a short-circuit evaluation is not a ternary operator, but often the ternary is used in cases where the short circuit would be enough. It has a better readability and can be chained.
Simulating the Python ternary operator.
For example
a, b, x, y = 1, 2, 'a greather than b', 'b greater than a'
result = (lambda:y, lambda:x)[a > b]()
Output:
'b greater than a'
a if condition else b
Just memorize this pyramid if you have trouble remembering:
condition
if else
a b
The ternary conditional operator simply allows testing a condition in a single line replacing the multiline if-else making the code compact.
Syntax:
[on_true] if [expression] else [on_false]
1- Simple Method to use ternary operator:
# Program to demonstrate conditional operator
a, b = 10, 20
# Copy value of a in min if a < b else copy b
min = a if a < b else b
print(min) # Output: 10
2- Direct Method of using tuples, Dictionary, and lambda:
# Python program to demonstrate ternary operator
a, b = 10, 20
# Use tuple for selecting an item
print( (b, a) [a < b] )
# Use Dictionary for selecting an item
print({True: a, False: b} [a < b])
# lambda is more efficient than above two methods
# because in lambda we are assure that
# only one expression will be evaluated unlike in
# tuple and Dictionary
print((lambda: b, lambda: a)[a < b]()) # in output you should see three 10
3- Ternary operator can be written as nested if-else:
# Python program to demonstrate nested ternary operator
a, b = 10, 20
print ("Both a and b are equal" if a == b else "a is greater than b"
if a > b else "b is greater than a")
Above approach can be written as:
# Python program to demonstrate nested ternary operator
a, b = 10, 20
if a != b:
if a > b:
print("a is greater than b")
else:
print("b is greater than a")
else:
print("Both a and b are equal")
# Output: b is greater than a
Vinko Vrsalovic's answer is good enough. There is only one more thing:
Note that conditionals are an expression, not a statement. This means you can't use assignment statements or pass or other statements within a conditional expression
Walrus operator in Python 3.8
After the walrus operator was introduced in Python 3.8, something changed.
(a := 3) if True else (b := 5)
gives a = 3 and b is not defined,
(a := 3) if False else (b := 5)
gives a is not defined and b = 5, and
c = (a := 3) if False else (b := 5)
gives c = 5, a is not defined and b = 5.
Even if this may be ugly, assignments can be done inside conditional expressions after Python 3.8. Anyway, it is still better to use normal if statement instead in this case.
More a tip than an answer (I don't need to repeat the obvious for the hundredth time), but I sometimes use it as a one-liner shortcut in such constructs:
if conditionX:
print('yes')
else:
print('nah')
, becomes:
print('yes') if conditionX else print('nah')
Some (many :) may frown upon it as unpythonic (even, Ruby-ish :), but I personally find it more natural - i.e., how you'd express it normally, plus a bit more visually appealing in large blocks of code.
You can do this:
[condition] and [expression_1] or [expression_2];
Example:
print(number%2 and "odd" or "even")
This would print "odd" if the number is odd or "even" if the number is even.
The result: If condition is true, exp_1 is executed, else exp_2 is executed.
Note: 0, None, False, emptylist, and emptyString evaluates as False.
And any data other than 0 evaluates to True.
Here's how it works:
If the condition [condition] becomes "True", then expression_1 will be evaluated, but not expression_2.
If we "and" something with 0 (zero), the result will always to be false. So in the below statement,
0 and exp
The expression exp won't be evaluated at all since "and" with 0 will always evaluate to zero and there is no need to evaluate the expression. This is how the compiler itself works, in all languages.
In
1 or exp
the expression exp won't be evaluated at all since "or" with 1 will always be 1. So it won't bother to evaluate the expression exp since the result will be 1 anyway (compiler optimization methods).
But in case of
True and exp1 or exp2
The second expression exp2 won't be evaluated since True and exp1 would be True when exp1 isn't false.
Similarly in
False and exp1 or exp2
The expression exp1 won't be evaluated since False is equivalent to writing 0 and doing "and" with 0 would be 0 itself, but after exp1 since "or" is used, it will evaluate the expression exp2 after "or".
Note:- This kind of branching using "or" and "and" can only be used when the expression_1 doesn't have a Truth value of False (or 0 or None or emptylist [ ] or emptystring ' '.) since if expression_1 becomes False, then the expression_2 will be evaluated because of the presence "or" between exp_1 and exp_2.
In case you still want to make it work for all the cases regardless of what exp_1 and exp_2 truth values are, do this:
[condition] and ([expression_1] or 1) or [expression_2];
Many programming languages derived from C usually have the following syntax of the ternary conditional operator:
<condition> ? <expression1> : <expression2>
At first, the Python's benevolent dictator for life (I mean Guido van Rossum, of course) rejected it (as non-Pythonic style), since it's quite hard to understand for people not used to C language. Also, the colon sign : already has many uses in Python. After PEP 308 was approved, Python finally received its own shortcut conditional expression (what we use now):
<expression1> if <condition> else <expression2>
So, firstly it evaluates the condition. If it returns True, expression1 will be evaluated to give the result, otherwise expression2 will be evaluated. Due to lazy evaluation mechanics – only one expression will be executed.
Here are some examples (conditions will be evaluated from left to right):
pressure = 10
print('High' if pressure < 20 else 'Critical')
# Result is 'High'
Ternary operators can be chained in series:
pressure = 5
print('Normal' if pressure < 10 else 'High' if pressure < 20 else 'Critical')
# Result is 'Normal'
The following one is the same as previous one:
pressure = 5
if pressure < 20:
if pressure < 10:
print('Normal')
else:
print('High')
else:
print('Critical')
# Result is 'Normal'
Yes, Python have a ternary operator, here is the syntax and an example code to demonstrate the same :)
#[On true] if [expression] else[On false]
# if the expression evaluates to true then it will pass On true otherwise On false
a = input("Enter the First Number ")
b = input("Enter the Second Number ")
print("A is Bigger") if a>b else print("B is Bigger")
Other answers correctly talk about the Python ternary operator. I would like to complement by mentioning a scenario for which the ternary operator is often used, but for which there is a better idiom. This is the scenario of using a default value.
Suppose we want to use option_value with a default value if it is not set:
run_algorithm(option_value if option_value is not None else 10)
or, if option_value is never set to a falsy value (0, "", etc.), simply
run_algorithm(option_value if option_value else 10)
However, in this case an ever better solution is simply to write
run_algorithm(option_value or 10)
The syntax for the ternary operator in Python is:
[on_true] if [expression] else [on_false]
Using that syntax, here is how we would rewrite the code above using Python’s ternary operator:
game_type = 'home'
shirt = 'white' if game_type == 'home' else 'green'
It's still pretty clear, but much shorter. Note that the expression could be any type of expression, including a function call, that returns a value that evaluates to True or False.
Python has a ternary form for assignments; however there may be even a shorter form that people should be aware of.
It's very common to need to assign to a variable one value or another depending on a condition.
>>> li1 = None
>>> li2 = [1, 2, 3]
>>>
>>> if li1:
... a = li1
... else:
... a = li2
...
>>> a
[1, 2, 3]
^ This is the long form for doing such assignments.
Below is the ternary form. But this isn't the most succinct way - see the last example.
>>> a = li1 if li1 else li2
>>>
>>> a
[1, 2, 3]
>>>
With Python, you can simply use or for alternative assignments.
>>> a = li1 or li2
>>>
>>> a
[1, 2, 3]
>>>
The above works since li1 is None and the interpreter treats that as False in logic expressions. The interpreter then moves on and evaluates the second expression, which is not None and it's not an empty list - so it gets assigned to a.
This also works with empty lists. For instance, if you want to assign a whichever list has items.
>>> li1 = []
>>> li2 = [1, 2, 3]
>>>
>>> a = li1 or li2
>>>
>>> a
[1, 2, 3]
>>>
Knowing this, you can simply such assignments whenever you encounter them. This also works with strings and other iterables. You could assign a whichever string isn't empty.
>>> s1 = ''
>>> s2 = 'hello world'
>>>
>>> a = s1 or s2
>>>
>>> a
'hello world'
>>>
I always liked the C ternary syntax, but Python takes it a step further!
I understand that some may say this isn't a good stylistic choice, because it relies on mechanics that aren't immediately apparent to all developers. I personally disagree with that viewpoint. Python is a syntax-rich language with lots of idiomatic tricks that aren't immediately apparent to the dabbler. But the more you learn and understand the mechanics of the underlying system, the more you appreciate it.
Pythonic way of doing the things:
"true" if var else "false"
But there always exists a different way of doing a ternary condition too:
"true" and var or "false"
There are multiple ways. The simplest one is to use the condition inside the "print" method.
You can use
print("Twenty" if number == 20 else "Not twenty")
Which is equivalent to:
if number == 20:
print("Twenty")
else:
print("Not twenty")
In this way, more than two statements are also possible to print. For example:
if number == 20:
print("Twenty")
elif number < 20:
print("Lesser")
elif 30 > number > 20:
print("Between")
else:
print("Greater")
can be written as:
print("Twenty" if number == 20 else "Lesser" if number < 20 else "Between" if 30 > number > 20 else "Greater")
The if else-if version can be written as:
sample_set="train" if "Train" in full_path else ("test" if "Test" in full_path else "validation")
Yes, it has, but it's different from C-syntax-like programming languages (which is condition ? value_if_true : value_if_false
In Python, it goes like this: value_if_true if condition else value_if_false
Example: even_or_odd = "even" if x % 2 == 0 else "odd"
A neat way to chain multiple operators:
f = lambda x,y: 'greater' if x > y else 'less' if y > x else 'equal'
array = [(0,0),(0,1),(1,0),(1,1)]
for a in array:
x, y = a[0], a[1]
print(f(x,y))
# Output is:
# equal,
# less,
# greater,
# equal
I find the default Python syntax val = a if cond else b cumbersome, so sometimes I do this:
iif = lambda (cond, a, b): a if cond else b
# So I can then use it like:
val = iif(cond, a, b)
Of course, it has the downside of always evaluating both sides (a and b), but the syntax is way clearer to me.